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Advocard [28]
3 years ago
6

Dianne's teacher has a chunk of dry ice for the class to look at. Dry ice is a kind of frozen gas—it is very cold. The teacher w

arns her class not to touch it. When Dianne holds her hand near it, it feels like the dry ice is radiating cold. What is really happening?
A) Heat energy from Dianne's hand is being transferred to the dry ice even though she is not touching it.
B) Chemicals in the dry ice are tricking Diane's nerves into thinking she feels cold.
C) Cold energy from the dry ice is being transferred to Diane's hand even though she is not touching it.
D) Small pieces of the dry ice are breaking off and hitting Diane's hand, making it feel cold.
Physics
2 answers:
Gelneren [198K]3 years ago
4 0
The answer is A. Hope this helps you with your work.
Lera25 [3.4K]3 years ago
3 0

Answer:

A) Heat energy from Dianne's hand is being transferred to the dry ice even though she is not touching it.

Explanation:

In this example, we learn that Dianne feels cold when she holds her hand close to the dry ice. In fact, Dianne feels as it the cold is moving from the dry ice to her hand. However, this is not what is happening. Changes in temperature occur when heat is transferred from a warm body to a colder one. Therefore, what is actually occurring is that heat energy from Dianne's hand is being transferred to the dry ice even though she is not touching it, and this makes her feel cold.

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In The United States how much more sugar is the average person consume each year that in 1970?
zmey [24]

In the 1970, the average American ate only 2 pounds of sugar a year. In 1970, we ate 123 pounds of sugar per year. Today, the average American consumes almost 152 pounds of sugar in one year. This is equal to 3 pounds (or 6 cups) of sugar consumed in one week!

7 0
3 years ago
When the temperature of 2.35 m^3 of a liquid is increased by 48.5 degrees Celsius, it expands by 0.0920 m^3. What is its coeffic
alexandr1967 [171]

Coefficient of volume expansion is 8.1 ×10⁻⁴ C⁻¹.

<u>Explanation:</u>

The volume expansion of a liquid is given by ΔV,

ΔV = α V₀ ΔT

ΔT = change in temperature  = 48.5° C

α =  coefficient of volume expansion =?

V₀ = initial volume = 2.35 m³

We need to find α , by plugin the given values in the equation and by rearranging the equation as,

\alpha=\frac{\Delta \mathrm{V}}{\mathrm{V}_{0} \Delta \mathrm{T}}=\frac{0.0920}{2.35 \times 48.5}=0.00081

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5 0
3 years ago
A transverse sinusoidal wave on a string has a period T = 25.0 ms and travels in the negative x direction with a speed of 30.0 m
lesya [120]

Answer:

a) A =0.021525m

b) \phi=0.37869rad

c) v_{max}=5.4098\frac{m}{s}

d)y(x,t)=(0.021525m)cos(\frac{8\pi}{3}x+80\pi t+0.37869)

Explanation:

1) Notation

A= Amplitude

v= velocity

\lambda= wavelength

k= wave number

\omega= angular frequency

f= frequency

2) Part a and b

The equation of movement for a transverse sinusoidal wave is gyben by (1)

y(t)=Acos(kx+ \omega t +\phi)   (1)

At x=0 ,t=0 we have that:

0.02=Acos(\phi)

The velocity would be the derivate of the position, so taking the derivate of (1) respect to t we got (2)

v(t)=-\omega Asin(kx+ \omega t+\phi)   (2)

And replacing the conditions at x=0, t=0 we got

-2\frac{m}{s}=-\omega Asin(\phi)  

Now we can find the angular frequency with equation (3)

\omega =\frac{2\pi}{T}   (3)

Replacing the values obtained we got:

\omega =\frac{2\pi}{0.025s}=80\pi \frac{rad}{s}  

From equation (1) we have:

Acos(\phi)=0.02   (a)

-2=-80\pi Asin(\phi)   (b)

So from condition (b) we have:

Asin(\phi)=\frac{1}{40\pi}   (c)

If we divide condition (c) by condition (a) we got:

\frac{Asin(\phi)}{Acos(\phi)}=tan(\phi)=\frac{1}{0.02x40\pi}=\frac{1}{0.8\pi}=0.39789

If we solve for \phi we got:

\phi =tan^{-1}(0.39789)=0.37869

And now since we have \phi we can find A from equation (a)

Acos(0.37869)=0.02

So then Solving for A we got A=\frac{0.02}{cos(0.37869)}=0.021525

3) Part c

From equation (2) we can see that the maximum speed occurs when sin(\omega t+\phi)=1, so on this case we have:

v_{max}=\omega A=80\pi \frac{rad}{s}x0.021525m=5.4098\frac{m}{s}

4) Part d

On this case we need an equation like (1), and we have everything except the wave number, and we can obtain this from the following expression:

v=\lambda f=\frac{2\pi}{k}\frac{\omega}{2\pi}=\frac{\omega}{k}   (4)

And solving for k from equation (4) we got

k=\frac{\omega}{v}=\frac{80\pi \frac{rad}{s}}{30\frac{m}{s}}=\frac{8\pi}{3}m^{-1}}

And with the k number we have everythin in order to create the wave function, given by:

y(x,t)=(0.021525m)cos(\frac{8\pi}{3}x+80\pi t+0.37869)

7 0
3 years ago
A tennis pro charges $15 per hour for tennis lessons for children and $30 per hour for tennis lessons for adults. The tennis pro
AURORKA [14]

Answer:

Third-degree price discrimination

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Atomic mass = protons + neutrons

So the total number of protons and neutrons
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3 years ago
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